Question

Electron in Hydrogen atom first jumps from third excited state to second excited state and then from second excited state to first excited state. The ratio of the wavelengths \( \lambda_1 : \lambda_2 \) emitted in the two cases respectively is

• A. \( \frac{7}{5} \)
• B. \( \frac{27}{20} \)
• C. \( \frac{27}{5} \)
• D. \( \frac{20}{7} \)

Hint:

When dealing with transitions between energy levels in hydrogen, the wavelength of emitted light is inversely proportional to the energy difference between the levels.

Answer 1

The Correct Option is D

Step 1: Understanding the problem.
The electron in the hydrogen atom transitions between different energy levels. The wavelengths emitted during these transitions are related to the energy differences between the levels. The energy of a photon emitted when the electron transitions between energy levels is given by: \[ E = \frac{13.6}{n^2} \quad \text{(in eV)} \] where \( n \) is the principal quantum number. The wavelength \( \lambda \) of the emitted photon is related to the energy by: \[ \lambda = \frac{hc}{E} \] where \( h \) is Planck’s constant, and \( c \) is the speed of light.
Step 2: Applying the energy level formula.
The transition from the third to the second excited state (i.e., \( n_3 = 3 \) to \( n_2 = 2 \)) gives the first wavelength \( \lambda_1 \), and the transition from the second to the first excited state (i.e., \( n_2 = 2 \) to \( n_1 = 1 \)) gives the second wavelength \( \lambda_2 \). The ratio of the wavelengths is given by: \[ \frac{\lambda_1}{\lambda_2} = \frac{E_2}{E_1} \] By calculating the energy differences between the two transitions, we get: \[ \frac{\lambda_1}{\lambda_2} = \frac{20}{7} \]
Step 3: Conclusion.
The ratio of the wavelengths is \( \frac{20}{7} \), which corresponds to option (D).